# Ergodicity of induced system

Suppose $$(X,\mathcal{F},\mu,T)$$ is an ergodic measure preserving dynamical system. Let $$Y\subset X$$ be such that $$\mu(Y)>0$$ and suppose there is an integrable function $$R:Y\to \mathbb{N}$$ such that $$T^{R(y)}(y)\in Y$$.

Then we can define a function $$F:Y\to Y$$ by $$F=T^R$$ and consider the induced system $$(Y,\mathcal{F}\cap Y, \mu|_Y,F)$$.

Can we say that $$F$$ is ergodic?

When $$R(x)=\inf\{n\ge 1 : T^n(x)\in Y\}$$, this induced system is very well studied and the answer to my question is yes (see for example http://www.weizmann.ac.il/math/sarigo/sites/math.sarigo/files/uploads/ergodicnotes.pdf, Theorem 1.7). However, the given proof does not extend to the general return times I am considering above.

Thanks :)

In general $$F$$ is not ergodic. A very simple example can be constructed as follows: let $$X=\mathbb{Z}_3=\{0,1,2\}$$ and $$\mu =1/3(\delta_0+\delta_1+\delta_2)$$ and $$T(x):=x+1$$. This is an ergodic system. Let us define $$Y:=\{0,1\}$$ and $$R\equiv 3$$. Since $$F=T^3=id$$, it is not ergodic.

• Is there any counter example with $T$ totally ergodic? Mar 23, 2020 at 20:39
• Isn't $T$ totally ergodic in Alejandro's counter example?
– Joël
Mar 24, 2020 at 0:45
• @Joël, Doesn't totally ergodic mean that all iterates are ergodic? That certainly is not the case if $T^3$ is identity. Mar 24, 2020 at 1:45
• Totally ergodic means that there exists only one $T$-invariant measure (it is a property of the space, the $\sigma$-algebra, and $T$, but not of the measure $\mu$)
– Joël
Mar 24, 2020 at 2:06
• @Joël I think that's unique ergodicity: en.wikipedia.org/wiki/Ergodicity#Unique_ergodicity Mar 24, 2020 at 2:22

I don't think that this system is even automatically measure-preserving (unlike the traditional induced map, which as you note is measure-preserving and ergodic).

Just take something silly like $$(X, T)$$ an irrational circle rotation with Haar measure, $$Y$$ the left half $$[0, 1/2)$$, and $$R$$ the first return time of a point in $$Y$$ to $$[0, 1/4)$$ (this always exists by minimality).

Then $$T^R(y) \in [0, 1/4))$$ for all $$y \in [0, 1/2)$$, so, for instance, $$\mu|_Y([0, 1/4)) = 1/2$$, but $$\mu|_Y\big((T^R)^{-1}([0, 1/4))\big) = \mu|_Y(Y) = 1$$. (I am here normalizing $$\mu|_Y$$ to make it a probability measure.)

I know my example is silly in that it's not even surjective on $$Y$$, but this isn't the problem; you could make a slightly trickier example where $$R$$ is the first return time to $$[0, 1/4)$$ for $$y \in [0, 3/8)$$ and the first return time to $$[1/4, 1/2)$$ for $$y \in [3/8, 1/2)$$, and then $$T^R$$ is surjective on $$Y$$, but still not measure-preserving.